## anonymous 4 years ago prove by induction that 1^2 + 2^2 + 3^2 ... + n^2 = n(n+1)(2n+1)/6

1)for n=1$1=\frac{1(1+1)(2+1)}6=1\checkmark$2) assuming it's true for some n, we just write out the formula$\sum_{i=1}^{n} i=\frac{n(n+1)(2n+1)}6$3) now we try to show it for n+1$\sum_{i=1}^{n+1} i=1+2+3+\dots+n^2+(n+1)^2$$=\frac{n(n+1)(2n+1)}6+(n+1)^2$$=\frac{n(n+1)(2n+1)+6(n+1)^2}6$$=\frac{(n+1)[n(2n+1)+6(n+1)]}6$$=\frac{(n+1)(2n^2+n+6n+6)}6$$=\frac{(n+1)(2n^2+7n+6)}6$$=\frac{(n+1)(n+2)(2n+3)}6$$=\frac{(n+1)(n+2)(2(n+1)+1)}6$which is the same formula replacing n with n+1$\text{QED}$